Why Analyze the Branches Separately?
A vertical asymptote divides a graph into separate branches. The curve may bend upward on one side and downward on the other, or it may bend the same way on both sides. A single concavity description that ignores the division can hide this difference.
The practical method is to find the domain, split it at the asymptote, and use the second derivative to determine concavity on every resulting interval. A sign chart makes the conclusions visible, while also preventing a common mistake: treating the asymptote as if the graph passed through it.
Second-Derivative Sign Charts Near an Asymptote
On an interval where \(f\) is twice differentiable, \(f''(x)>0\) means that \(f\) is concave up there, and \(f''(x)<0\) means that it is concave down. To build a sign chart near \(x=c\), include \(c\) as a boundary even though it is not in the interval being tested. Also include any zeros or other points where \(f''\) is undefined.
The resulting intervals must remain separate across \(c\). For instance, a chart may have the intervals \((-\infty,c)\) and \((c,\infty)\). If \(f''\) has a zero on one of those branches, that zero splits that branch into more intervals; it does not remove the division at the asymptote.
Identify \(x=c\) and determine which intervals belong to the domain. A vertical asymptote may exist even if a finite value has been assigned to \(f(c)\).
Factor or simplify \(f''(x)\) when possible. Mark its zeros and points where it is undefined, as well as the asymptote.
Use a test value in each interval, or determine the sign from the factors. Record the sign of \(f''\), then state the concavity.
A change of concavity across a point does not make the asymptote an inflection point. Under the usual AP convention, an inflection point is a point on the graph where the function is continuous and its concavity changes.
Why the Sign of \(f''\) Determines the Bending
The second derivative measures how the first derivative changes. On an interval where \(f''>0\), the first derivative increases, so tangent slopes grow as \(x\) increases: the graph bends upward. On an interval where \(f''<0\), tangent slopes decrease, so the graph bends downward.
This connection can be justified directly. Suppose \(f''(x)>0\) throughout an interval and choose any \(a<b\) in that interval. By the Mean Value Theorem applied to \(f'\), there is a number \(\xi\) between \(a\) and \(b\) such that
Because \(f''(\xi)>0\) and \(b-a>0\), it follows that \(f'(b)>f'(a)\). Thus the tangent slopes increase throughout the interval, which is the second-derivative test for concavity up. If \(f''<0\), the same equation gives \(f'(b)<f'(a)\), so the tangent slopes decrease and the graph is concave down. This reasoning applies on each interval separately; it does not require the function to be defined across a vertical asymptote.
Worked Example: Opposite Concavity on the Two Branches
Worked Example: Opposite Concavity on the Two Branches
Analyze the concavity of \(f(x)=\frac{1}{x}\), including the behavior near its vertical asymptote.
The domain is all real numbers except \(x=0\). The one-sided limits are
Therefore \(x=0\) is a vertical asymptote, and the domain has two branches: \((-\infty,0)\) and \((0,\infty)\).
Differentiate twice:
For \(x<0\), \(x^3<0\), so \(f''(x)<0\). For \(x>0\), \(x^3>0\), so \(f''(x)>0\). A sign chart records the result:
| Interval | Test value | Sign of \(f''\) | Concavity |
|---|---|---|---|
| \((-\infty,0)\) | \(-1\) | \(2/(-1)^3=-2<0\) | Down |
| \((0,\infty)\) | \(1\) | \(2/1^3=2>0\) | Up |
The left branch is concave down, and the right branch is concave up. The derivative check is consistent: differentiating \(f'(x)=-x^{-2}\) gives \(f''(x)=2x^{-3}\), as used in the chart. Although the concavity differs on the two sides of zero, \(x=0\) is not an inflection point: the function is not continuous there, and the point is not on its graph.
Worked Example: Matching Concavity on Both Branches
Worked Example: Matching Concavity on Both Branches
Analyze \(g(x)=\frac{1}{(x-2)^2}\) on either side of its vertical asymptote.
The function is undefined at \(x=2\), and
Thus \(x=2\) is a vertical asymptote, and the domain intervals are \((-\infty,2)\) and \((2,\infty)\). Write \(g(x)=(x-2)^{-2}\). Differentiating gives
For every \(x\ne2\), the denominator \((x-2)^4\) is positive, so \(g''(x)>0\) on both intervals:
| Interval | Test value | Sign of \(g''\) | Concavity |
|---|---|---|---|
| \((-\infty,2)\) | \(1\) | \(6/(1-2)^4=6>0\) | Up |
| \((2,\infty)\) | \(3\) | \(6/(3-2)^4=6>0\) | Up |
Both branches are concave up, even though each is separated from the other by the asymptote. Checking the derivative, differentiating \(-2(x-2)^{-3}\) gives \((-2)(-3)(x-2)^{-4}=6(x-2)^{-4}\), confirming the second derivative and its positive sign on both branches.
Worked Example: An Inflection Point on One Branch
Worked Example: An Inflection Point on One Branch
Analyze the concavity of \(h(x)=x^2+\frac{1}{x-1}\). Determine whether any inflection point occurs.
The function is undefined at \(x=1\). Since the reciprocal term tends to \(-\infty\) from the left and \(+\infty\) from the right, \(x=1\) is a vertical asymptote. The branches are \((-\infty,1)\) and \((1,\infty)\).
Differentiate each term:
To find where the second derivative is zero, solve:
For \(x\ne1\), combine the terms in \(h''\) and factor:
The quadratic factor is always positive because
Therefore the sign of \(h''\) is determined by \(x/(x-1)^3\). Test the intervals separated by \(x=0\) and \(x=1\):
| Interval | Test value | Sign of \(h''\) | Concavity |
|---|---|---|---|
| \((-\infty,0)\) | \(-1\) | \((-1)/(-2)^3>0\) | Up |
| \((0,1)\) | \(1/2\) | \((1/2)/(-1/2)^3<0\) | Down |
| \((1,\infty)\) | \(2\) | \(2/1^3>0\) | Up |
Concavity changes at \(x=0\), and \(h\) is continuous there. Its value is
Thus \((0,-1)\) is an inflection point. There is no inflection point at \(x=1\): the function is not defined there, and the branches are separated by a vertical asymptote. As a derivative check, differentiating \(h'(x)=2x-(x-1)^{-2}\) gives \(h''(x)=2+2(x-1)^{-3}\), agreeing with the expression used in the sign chart.
A Common Mistake: Calling the Asymptote an Inflection Point
A tempting but incorrect argument is: “For \(f(x)=1/x\), the graph is concave down to the left of zero and concave up to the right, so \((0,\text{undefined})\) is an inflection point.” This confuses a change between two branches with a point on the graph. There is no graph point at \(x=0\), and the function is discontinuous there.
The correction is to report the two concavity intervals separately: concave down on \((-\infty,0)\) and concave up on \((0,\infty)\). Do not label the asymptote as an inflection point. More generally, the function may be assigned a finite value at the asymptote without removing it. For example, assigning \(f(0)=7\) while keeping \(f(x)=1/x\) for \(x\ne0\) leaves the one-sided limits infinite. The function is still discontinuous at zero, so under the usual AP convention zero is not an inflection point.
What the Branch-by-Branch Analysis Tells You
Concavity analysis near a vertical asymptote describes the bending of each piece of the graph, not a transition through the asymptote. The two pieces may have opposite concavity, as for \(1/x\); matching concavity, as for \(1/(x-2)^2\); or additional changes within a branch, as in the example with \(x^2+1/(x-1)\).
For an accurate analysis, make sure each interval in the sign chart lies entirely in the domain and on one side of the asymptote. A zero of \(f''\) may mark a potential change of concavity on a branch, but it is an inflection point only if the function is continuous there and the concavity actually changes. The asymptote itself is a boundary between branches, not a point where the graph changes concavity.
Check Your Understanding
Use the second derivative and keep the intervals on either side of each vertical asymptote distinct.
- For \(p(x)=1/(x+3)\), find the vertical asymptote and state the concavity on each branch.
- For \(q(x)=1/(x+1)^4\), compute \(q''(x)\) and determine the concavity on both sides of \(x=-1\).
- For \(r(x)=x^2+1/(x-2)\), find the zeros of \(r''\) and make a concavity sign chart.
- Why is a change from concave down to concave up across a vertical asymptote not, by itself, an inflection point?
- If a finite value is assigned to a function at \(x=c\), can \(x=c\) still be a vertical asymptote? Explain using the one-sided-limit criterion.
